Do any of you happen to know the history of the standard prime factorization proof of $\sqrt 2$ is irrational? I know this theorem was known to Aristotle, and that the Fundamental Theorem of Arithmetic, on which the proof rests, is found already in Euclid, but I've not been able to track down the origin of this particular proof.

3 Answers
3

"The Discovery of Incommensurability" by Kurt von Fritz [ http://www.jstor.org/stable/1969021 ] indicates that the early Greek mathematicians did not explicitly use the Fundamental Theorem to prove the irrationality of √2. The proof known to Aristotle ("the diagonal of the square is incommensurate with the side, because odd numbers are equal to evens if it is supposed to be commensurate") uses a restricted version of the Fundamental Theorem, as explained in http://en.wikipedia.org/wiki/Quadratic_irrational

Apparently, the explicit use of the Fundamental Theorem to prove the irrationality of √2 is post-Gauss. This is argued convincingly by Barry Mazur:

This fundamental theorem of arithmetic has a peculiar history. It is not trivial, and any of its proofs take work, and, indeed, are interesting in themselves. But it is nowhere stated in the ancient literature. It was used, implicitly, by the early modern mathematicians, Euler included, without anyone noticing that it actually required some verification, until Gauss finally realized the need for stating it explicitly, and proving it.

Thanks for the Mazur paper. It is very good! But is it so clear that Gauss applied the fundamental theorem to the $\sqrt 2$. Or am I missing something?
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ManyaSep 18 '12 at 8:47

well, once you have proven the fundamental theorem, there does not seem to be anything left to prove regarding the irrationality of the square root of any number that is not a perfect square; I imagine asking Gauss why he doesn't care to publish that proof and receiving a gentle smile..
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Carlo BeenakkerSep 18 '12 at 10:31

The Evolution of the Euclidean
Elements: A Study of the Theory of
Incommensurable Magnitudes and Its
Significance for Early Greek
Geometry (Dordrecht: D. Reidel Publishing Co., 1975).

This work incorporates Knorr's Ph.D.
thesis. It traces the early history of
irrational numbers from their first
discovery (in Thebes between 430 and
410 BC, Knorr speculates), through the
work of Theodorus of Cyrene, who
showed the irrationality of the square
roots of the integers up to 17, and
Theodorus' student Theaetetus, who
showed that all non-square integers
have irrational square roots. Knorr
reconstructs an argument based on
Pythagorean triples and parity that
matches the story in Plato's
Theaetetus of Theodorus' difficulties
with the number 17, and shows that
switching from parity to a different
dichotomy in terms of whether a number
is square or not was the key to
Theaetetus' success.

Here is a proof based on the well-ordering of the positive integers rather than the FTA:

To begin with, we observe that if $\sqrt{2}$ is rational, then there is some positive integer q such that q × $\sqrt{2}$ is an integer. Since the positive integers are well ordered, we may suppose that q is the smallest such number.
We next observe that since 1 < $\sqrt{2}$ < 2, then $\sqrt{2}$ – 1 < 1, and consequently q × ($\sqrt{2}$ – 1) = (q × $\sqrt{2}$ – q ) is less than q. Let us call this new number r, and observe that it too is a positive integer. But we now have r × $\sqrt{2}$ is also an integer, since r × $\sqrt{2}$ = (q ×$\sqrt{2}$ – q ) × $\sqrt{2}$ = (2q – q × $\sqrt{2}$). In short, r is a positive integer less than q and r × $\sqrt{2}$ is an integer. But we said that q was the smallest positive integer with this property, and so we have a contradiction.

A variation on this: If $\sqrt{2}$ is a rational number, say $a/b$ with $a,b\in \mathbb{Z}$, then $\sqrt{2}-1$ is a rational between 0 and 1. Hence $(\sqrt{2}-1)^n\to 0$ as $n\to\infty$. But $(\sqrt{2}-1)^n$ has the form $u\frac{a}{b}+v$ for some $u,v\in\mathbb{Z}$, so $(\sqrt{2}-1)^n\ge1/b$. Contradiction.
–
SJRSep 15 '12 at 4:58

8

I fail to see what this has to do with the question.
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Franz LemmermeyerSep 15 '12 at 13:00

Your right Franz, it doesn't. It's just that there seems to be a belief that you NEED unique prime factorization to prove the irrationality of non-square integers, and when I first saw this (much more elementary) proof I found it an eye-opening experience.
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Dick PalaisSep 15 '12 at 14:56

1

This is a nice proof, but I don't think the other standard proof (write $\sqrt{2} = a/b$ with $a,b$ not both even) requires unique factorization. It only requires one lemma about even and odd squares.
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Ryan ReichSep 16 '12 at 0:31

1

Yeah, I agree, this thread is off-topic, but there are of course other cool proofs out there. You can find a nice collection at the link I gave above (cut-the-knot.org/proofs/sq_root.shtml). If you haven't seen 8''' before, that one is quite nice.
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ManyaSep 18 '12 at 11:32